Could a packed fractional still of the same type used to distill ethanol to 95% be used to produce deuterium depleted drinking water? The still has a 1000 watt electric heating element and has a fractioning column that is 120cm long which is packed with stainless steel wool. If it is feasible what kind of performance might I expect?

  • $\begingroup$ Some people hope to live longer by drinking it. Well, as to me, this is harmless, so why not. As for the technical part, we had a similar question not so long ago: chemistry.stackexchange.com/questions/47189/…. Distillation in this setup is hardly going to do the trick; after all, water and heavy water are much more similar than water and ethanol. $\endgroup$ Mar 12, 2016 at 14:29
  • $\begingroup$ Consuming deuterium depleted water may have health benefits. A company in Europe sells DDW and it is very expensive at around $40 per liter. $\endgroup$
    – L Pottle
    Mar 12, 2016 at 14:30
  • $\begingroup$ I very much doubt the theories that deuterium depleted water provides measurable health benefits, but it would be a very expensive idea to test directly. Has anyone ever shown that providing mice with a diet of fivefold or even one hundred-fold more deuterium than usual results in altered health in any way? $\endgroup$
    – Curt F.
    Mar 13, 2016 at 3:07

1 Answer 1


Relying on the difference in vapor pressures of $\ce{HDO}$ and $\ce{H2O}$ is equivalent on relying on the equilibration of

$\ce{HDO(g) + H2O(l) <=> H2O(g) + HDO(l)}$

The fractionation factor $\alpha_{\ce{D}(l)/\ce{D}(g)}$ for this reaction is 1.015 at STP.

Deuterium ($\ce{^2H}$ or $\ce{D}$) is separated from protium ($\ce{^1H}$) commercially, usually via the Girdler sulfide process, which instead of relying on the above reaction, relies on

$\ce{ HDS + H2O <=> H2S + HDO}$

The fractionation factor $\alpha$ for this process is about 2.2.

Steam distillation for deuterium enrichment is certainly possible, but is not nearly as favorable as the industrial process. Given the very low concentration of deuterium in natural waters, the Rayleigh fractionation equation would be applicable:

$$R_{\ce{D}/\ce{H}} \approx R_{\ce{D}/\ce{H}_{init}} \times f^{\alpha-1}$$

In this equation:

  • $R_{\ce{D}/\ce{H}}$ is the ratio of deuterium to hydrogen in the liquid water
  • $R_{\ce{D}/\ce{H}_{init}}$ is the initial ratio of deuterium to hydrogen, or about 0.000156.
  • $f$ is the fraction of the liquid water remaining after distillation.
  • $\alpha$ is the fractionation factor mentioned above. However the value of 1.0015 is for the condensation of vapor phase water into liquid; performing the reverse reaction we should use the inverse, i.e. for distillation of existing liquid into vapor, $\alpha = 0.985$

With this equation and with mass balances (the total amount of deuterium and of protium doesn't change in the distillation process) you can find the amount of deuterium in the distillate.

If you distill off 100 grams of an initial 1000 grams of water, then $f=0.9$, $f^{\alpha-1} \approx 1.000156$, meaning the remaining 900 grams of liquid water will have a deuterium-to-hydrogen ratio of 0.0001562 instead of 0.000156.

The mass balance for deuterium would be:

$1000 \times 0.000156 = 100 R_{\ce{D}/{\ce{H}_{distillate}}} + 900 \times 0.0001562$

If I did all my math right, that means your distillate would have a deuterium-to-hydrogen ratio of 0.0001542, or about 1.2% less than the original water.

Hopefully this exercise convinces you that you would have to do a lot of distillation and burn through a ton of water to generate reasonable amounts of deuterium depleted water through simple steam distillation. There's a reason they don't do it that way commercially.

  • $\begingroup$ You mention "simple steam distillation" in your closing remarks. I completely agree with you on that idea. However, the method in fractional distillation is much more efficient as it performs many simple distillations (dozens or more re-distillations in a single pass) within the fractioning column apparatus. Would this not improve upon the 1.2% given in your formula? $\endgroup$
    – L Pottle
    Mar 16, 2016 at 21:41
  • 1
    $\begingroup$ You are right, bu.... Roughly speaking, say you get a 1.2% decrease in relative deuterium content for each equilibrium stage, assuming each stage produces 10% vapor and 90% liquid. Then you get an improvement factor $1.012^n$ for $n$ equilibrium stages, but the mass you obtain goes down by tenfold each time. So if you had three stages, you would get 3.6% less deuterium than you started with, but you would have only one gram of the resultant water. By optimizing the liquid/vapor ratio at each equilibrium stage, you could improve on this somewhat, but not by much. $\endgroup$
    – Curt F.
    Mar 16, 2016 at 22:04
  • $\begingroup$ Cody from Cody's Lab on YouTube used electrolysis and managed to concentrate an amount using time and solar electricity. - youtube.com/watch?v=VC4dk1JU4tQ $\endgroup$
    – KalleMP
    Feb 14, 2017 at 19:31

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